Problem: $\vec w = (6,-2)$ $\dfrac12\vec w= ($
Solution: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $\dfrac12 \vec{w}$ : $\begin{aligned} {\dfrac12}\vec w = {\dfrac12} \cdot (6,-2) &= \left({\dfrac12} \cdot 6, {\dfrac12} \cdot (-2)\right) \\\\ &= (3,-1) \end{aligned}$ The answer is $ (3,-1) $.